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Black_prince [1.1K]
3 years ago
8

If only one external force acts on a particle, does it necessarily change the particle's kinetic energy and velocity? (Select al

l that apply.) Particle's kinetic energy won't change. Particle's kinetic energy may or may not change. Particle's kinetic energy must change. Particle's velocity won't change. Particle's velocity may or may not change. Particle's velocity must change.
Physics
1 answer:
Rina8888 [55]3 years ago
4 0

Answer:

Option 3 and 6

Explanation:

If the particle is under the action of only one external force, both its kinetic energy and the velocity must vary.

When the particle is exerted by an external force, then this external force tends to accelerate the particle.

When the particle accelerates, its velocity must vary for that acceleration to occur.

When the velocity of the particle changes then its Kinetic energy must vary with the velocity of the particle.

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jonny [76]

Answer:

I DON'T SPEAK TACO BELL!

Explanation:

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3 years ago
Object A travels at a constant velocity of 2.0 m/s right, and object B travels in the same direction at a
pashok25 [27]

==> Object A travels for 60 seconds before Object B starts out.

==> Object A moves at 2 m/s.

==> So Object A has a lead of 120 m when Object B starts out.

==> Object B moves at 3 m/s . . . 1 m/s faster than Object A.

==> So Object B catches up on Object A by 1 m every second.

==> Object B closes up Object A's lead of 120 m in <em>120 seconds</em>.  

8 0
3 years ago
What is the relationship between mass, volume, and density? What is the relationship between mass, volume, and density? Density
kirza4 [7]

Answer:

A. Density is proportional to mass and inversely proportional to volume

Explanation:

Density can be defined as the property that matter has, whether solids, liquids or gases, to be compressed in a given space.

The expression that relates the density with the another values is given by,

\rho= \frac{m}{V}

Where,

\rho \propto m  (It is directly proportional to the mass of the object)

\rho = \frac{1}{V}(It is inversely proportional to the volumen of the object)

<em>Therefore density is directly proportional to the mass and inversely proportional to the volumen.</em>

4 0
3 years ago
when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is remo
Lera25 [3.4K]

Answer:

q1 = 7.6uC , -2.3 uC

q2 = 7.6uC , -2.3 uC

( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )

Explanation:

Solution:-

- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.

- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:

                              F = k\frac{|q_1|.|q_2|}{r^2}

Where,

                     k: The coulomb's constant = 8.99*10^9

- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.  

- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.

- Therefore, the force of attraction between the spheres would be:

                             \frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1 ... Eq 1

- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).

- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,

                             q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}

- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:

                          \frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2}  = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 =   \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6  .. Eq2

- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:

                         -\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123

                         

                          q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\

 

6 0
4 years ago
A moving walkway has a speed of 0.9 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of
Leona [35]
B or .2 East is the correct answer
3 0
3 years ago
Read 2 more answers
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